This type of foundation is the one that most resembles the foundations of traditional construction. In these cases, the vertical load is much more important than the horizontal one or the moment. The analysis of the foundation is practically reduced to find a contact surface that produces a stress on the ground smaller than the soil bearing capacity.
N / A ≤ sa
N = vertical load
A = footing horizontal surface
sa= soil bearing capacity
Note that in this case, unlike the case with a pulling force, we do not contemplate at all the friction of lateral sides of the footing. That’s because the foundation will not easily move. When we pull up, footing can move up, but when we push down the foundation does not move (settlement is not considered a movement that causes friction). In any case, if we continued to push the soil, it would break, locally or in a generalized manner, and we would get the value of the breaking load or stress. Generally, the value of the bearing capacity is obtained by applying a safety factor of 3 or 4 to the breaking stress.
Despite similarities, foundation of a vertical pushing load in a tensile structure has a series of particularities that must be pointed out:
- In traditional buildings, self weight and permanent loads are usually very important. Generally more important than live loads. In this case analysis of settlement is an important design factor. Many times it is the most important design factor. And not only by the magnitude of the load, but by its duration, which causes soil consolidation, a different settlement that must be added to the initial settlement (elastic).In tensile structures, most important loads are wind and snow. Both are live loads of short duration, and do not produce any type of soil consolidation.
- Soil bearing capacity is a value that depends on the data obtained through a geotechnical report. In these reports it is necessary to contemplate a sufficient depth of soil in order to foresee soft soil layers that may cause important settlements.
Settlements in traditional constructions can cause cracks or very important cracks, especially if there exist differential settlements. In tensile structures, settlements (if there are any) do not cause cracks or fissures. What they would cause is a slackening of the structure, which always has an easy repair, since it only needs to be re-stressed. In addition, differential settlements have no special meaning in a tense structure.
- Therefore, it will be necessary to consider that some bearing capacity values used for tensile structures are much (too much?) on the safety side. In fact, it is necessary to obtain the soil bearing capacity from the soil breaking load (generalized, local or punching breaking) and neglect the magnitude of settlements, mainly of the differential settlements.
As reference values we can use the ones given by the Spanish standard NBE-AE/88, already obsolete after the appearance of the Technical Building Code (Eurocode). It should be noted that this text indicates that these values do not guarantee a tolerable settlement for the building and that it is necessary to check it in a separate analysis. Nevertheless, we have already commented the much smaller impact of this settlement in tensile structures.
|Soil nature||Soil bearing capacity kg/cm², according to foundation depth in m|
|1 Rocks (unaltered, although some small crack is accepted)|
|2 Soils without cohesion (consolidated)*|
|3 Soils with cohesion|
|Very soft clay||–||–||0,5|
|4 Deficient soils|
|Organic soils||Normally null resistance, except if some value is defined experimentally.|
|Non consolidated soils|
a) The indicated values refer to consolidated soils that require the use of a peak to remove them. For medium consolidated soils in which the shovel penetrates with difficulty, the previous values will be multiplied by 0.8. For loose soils, which are easily removed with the shovel, the indicated values will be multiplied by 0.5
b) The indicated values correspond to a foundation width equal to or greater than 1m. lower widths, values will be multiplied by the width of the foundation in meters.
c) When the water table is at a distance from the resistant surface less than its width, the values in the table will be multiplied by 0.8.
In this case, we must refer to normal building types found in traditional building.
Since most of the time the element that transmits the vertical load is a metallic element (steel tube), it will be necessary to design a support plate on top of the foundation. This plate will be fixed to the foundation through some anchors or connectors. As the horizontal force is zero these anchors or connectors do not work practically.
The footing will have a reinforcement at the base, a rebar grid able to absorb tensile forces generated by the compression lines that go from the support plate to the base of the foundation. Normally, for reasons of durability, it is recommended
– to place a layer of blinding concrete at the base of the foundation
– a correct concrete cover of the grid (more than 5 cm)
– a minimum diameter of the rebars: 14 or 16 mm.
Actually, vertical loading examples are not very interesting. We only need to define a footing horizontal surface to produce a contact stress lower than the soil bearing capacity and to decide a depth of the foundation. In any case we should provide this minimum depth:
- 80 cm, depth at which we can consider that rain or frost no longer affect.
- Half the width of the foundation. A value that usually ensures that footing has a rigid element behaviour. (It is true, however, that in cases where soil is very deformable, it may be better to use a less deep footing with flexible behaviour, but that complicates the design much more and is beyond the scope of this text).
So, let’s suppose a case where:
- vertical load to be supported: 30 t
- soil: thick sand
As a reference value, according to the Basic Standard NBE/AE-88, at a depth of one meter, this type of soil would have a bearing capacity of 3.2 kg/cm². Then, the surface of the footing will be equal to or greater than
A = Fv / sa = 30 / 32 = 0,938 m²
a = √ 0,625 = 0,97 m ≈ 1 m
h = 0,8 m ≥ (a / 2)
So, footing will have these dimensions: 1 x 1 x 0,8 m